Question

Question number 2 , 4 &5 on
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Answer 1

Answer on Question#56568 - Physics - Mechanics - Relativity

Solution:

(2) The moment of inertia of one rod is $I_r = \frac{1}{12} ML^2$ . The moment of inertia of the cross about the axis $z$ perpendicular to it and passing through it center is

$$I _ {z} = 2 I _ {r} = \frac {1}{6} M L ^ {2}$$

Due to the symmetry of this cross relatively to $z$ -axis, the moments of inertia of the cross about the axes perpendicular to $z$ -axis and passing through its center are all the same and equal to half the $I_z$ (there is a corresponding theorem). Thus the required moment of inertia is

$$I = \frac {1}{2} I _ {z} = \frac {1}{1 2} M L ^ {2}$$

(4) The direction of the force of friction is opposite to the direction of the velocity vector of the point on the sphere that touches the ground. The direction of this velocity (taking into account that $v = \omega r$ ) is

$$\boldsymbol {\tau} _ {v} = \frac {1}{\sqrt {2}} (\hat {t} - \hat {k})$$

Therefore, the direction of the force of friction is

$$\boldsymbol {l} _ {f} = - \boldsymbol {\tau} _ {v} = \frac {1}{\sqrt {2}} (\hat {k} - \hat {t})$$

The magnitude of the force of friction is

$$F _ {f} = m g \mu$$

Thus the frictional force vector is

$$\boldsymbol {F} _ {f} = F _ {f} \cdot \boldsymbol {l} _ {f} = \frac {m g \mu}{\sqrt {2}} (\hat {k} - \hat {\imath})$$

(5) The moment of inertia of the disk about its center is

$$I _ {d} = \frac {1}{2} M R ^ {2}$$

According to the parallel axis theorem the moment of inertia about the origin is

$$I = I _ {d} + M R ^ {2} = \frac {3}{2} M R ^ {2}$$

**Answer:**

(2) (A)

(4) (C)

(5) (C)

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